Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4π‘Ÿ/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let ∠ BOC = ΞΈ Now, AC = AO + OC H = r + r cos ΞΈ H = r (1 + cos ΞΈ) And, R = r sin ΞΈ We need to maximize volume of cone. Volume of the cone is V = 1/3 πœ‹π‘…^2 𝐻 V = 1/3 πœ‹π‘Ÿ^2 sin^2⁑"ΞΈ" r (1 + cos ΞΈ) V = 𝟏/πŸ‘ 𝝅𝒓^πŸ‘ γ€–π’”π’Šπ’γ€—^𝟐⁑"ΞΈ" (1 + cos ΞΈ) Finding 𝒅𝑽/π’…πœ½ 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 (𝑑[sin^2⁑〖θ γ€— (1 + cos⁑θ))/π‘‘πœƒ 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 [2 sin⁑〖"ΞΈ" cos⁑〖"ΞΈ" (1+cos⁑〖"ΞΈ" )+sin^2⁑〖"ΞΈ" (βˆ’sin⁑〖"ΞΈ" )γ€— γ€— γ€— γ€— γ€— ] 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 (2 sin⁑〖"ΞΈ" cos⁑〖"ΞΈ" (1+cos⁑〖"ΞΈ" )βˆ’sin^3⁑"ΞΈ" γ€— γ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 (2 sin⁑〖"ΞΈ" cos⁑〖"ΞΈ" (1+cos⁑〖"ΞΈ" )βˆ’sin^3⁑"ΞΈ" γ€— γ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (2 cos⁑〖"ΞΈ" +γ€–2 cos^2〗⁑〖"ΞΈ" βˆ’γ€–π’”π’Šπ’γ€—^𝟐⁑"ΞΈ" γ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (2 cos⁑〖"ΞΈ" +γ€–2 cos^2〗⁑〖"ΞΈ" βˆ’(πŸβˆ’γ€–π’„π’π’”γ€—^𝟐⁑"ΞΈ" γ€— γ€—)) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (πŸ‘ 〖𝒄𝒐𝒔〗^πŸβ‘γ€–"ΞΈ" +γ€–πŸ 𝒄𝒐𝒔〗⁑〖"ΞΈ" βˆ’πŸγ€— γ€— ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (πŸ‘ 〖𝒄𝒐𝒔〗^πŸβ‘γ€–"ΞΈ" +γ€–πŸ‘ π’„π’π’”γ€—β‘πœ½ γ€— βˆ’π’„π’π’”β‘πœ½ βˆ’πŸ) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (3 γ€–π‘π‘œπ‘  〗⁑"ΞΈ" (cosβ‘πœƒ+1)βˆ’1(cosβ‘πœƒ+1)) 𝑑𝑉/π‘‘πœƒ = 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (3 π‘π‘œπ‘ β‘γ€–"ΞΈ" βˆ’1γ€— )(π‘π‘œπ‘ β‘γ€–"ΞΈ" +1γ€— ) Putting 𝒅𝑽/π’…πœ½ = 0 1/3 πœ‹π‘Ÿ^3 sin ΞΈ (π‘π‘œπ‘ β‘γ€–"ΞΈ" +1γ€— ) (3 π‘π‘œπ‘ β‘γ€–"ΞΈ" βˆ’1γ€— ) = 0 𝐬𝐒𝐧 ΞΈ (cos ΞΈ + 1) (3 cos "ΞΈ" βˆ’ 1) = 0 𝐬𝐒𝐧 ΞΈ = 0 ΞΈ = 0Β° ΞΈ cannot be 0Β° for cone 𝐜𝐨𝐬 ΞΈ + 1 = 0 cos ΞΈ = βˆ’1 For cone, 0Β° < ΞΈ < 90Β° & cos ΞΈ is negative in 2nd & 3rd quadrant. So cos ΞΈ = βˆ’1 is not possible πŸ‘ 𝐜𝐨𝐬 ΞΈ βˆ’ 1 = 0 cos ΞΈ = 1/3 ΞΈ = cosβˆ’1 1/3 cos ΞΈ = 𝟏/πŸ‘ is possible So, 𝒄𝒐𝒔⁑"ΞΈ" = 𝟏/πŸ‘ Thus, H = r (1 + cos ΞΈ) H = r ("1 + " 1/3) H = πŸ’π’“/πŸ‘ Finding (𝒅^𝟐 𝑽)/(π’…πœ½^𝟐 ) 𝑑𝑉/π‘‘πœƒ = (1/3 πœ‹π‘Ÿ^3 sinβ‘πœƒ (3 cos2 πœƒ + 2 cosβ‘πœƒβˆ’ 1)) = 1/3 πœ‹π‘Ÿ^3 [cosβ‘πœƒ (3 cos^2β‘πœƒ+2 cosβ‘πœƒβˆ’1)+sinβ‘πœƒ (6 cosβ‘πœƒ (βˆ’π‘ π‘–π‘›)βˆ’2 sinβ‘πœƒ )] = 1/3 πœ‹π‘Ÿ^3 [3 cos^3β‘πœƒ+2 cos^2β‘πœƒβˆ’cosβ‘πœƒ+sinβ‘πœƒ (βˆ’6 cosβ‘πœƒ sinβ‘πœƒβˆ’2 sinβ‘πœƒ )] = 1/3 πœ‹π‘Ÿ^3 [3 cos^2β‘πœƒ+2 cos^2β‘πœƒβˆ’cosβ‘πœƒβˆ’6 sin^2β‘πœƒ cosβ‘πœƒβˆ’2 sin^2β‘πœƒ ] Now, 𝐜𝐨𝐬⁑𝜽=𝟏/πŸ‘ And sin2 "ΞΈ" = 1 βˆ’ cos2 "ΞΈ" = 1 βˆ’ (1/3)^2= 1 βˆ’ 1/9 = πŸ–/πŸ— Putting values in (𝒅^𝟐 𝑽)/(π’…πœ½^𝟐 ) (𝑑^2 𝑉)/(π‘‘πœƒ^2 ) = 1/3 πœ‹π‘Ÿ^3 [3(1/3)^3+2(1/3)^2βˆ’1/3βˆ’6(1/3)(8/9)βˆ’2(8/9)] = 1/3 πœ‹π‘Ÿ^3 [3(1/27)+2(1/9)βˆ’1/3βˆ’6(8/27)βˆ’2(8/9)] = 1/3 πœ‹π‘Ÿ^3 [1/9+ 2/9βˆ’1/3βˆ’16/9βˆ’16/9] = 1/3 πœ‹π‘Ÿ^3 [(βˆ’32)/9] = (βˆ’πŸ‘πŸπ…π’“^πŸ‘)/πŸπŸ• Thus, (𝑑^2 𝑉)/(π‘‘πœƒ^2 ) < 0 for 𝐜𝐨𝐬⁑𝜽=𝟏/πŸ‘ So, V is maximum for 𝐜𝐨𝐬⁑𝜽=𝟏/πŸ‘ Hence, altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is πŸ’π’“/πŸ‘ .

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.